How to get folder name, in which given file resides, from pathlib.path?
It looks like there is a parents
element that contains all the parent directories of a given path. E.g., if you start with:
>>> import pathlib>>> p = pathlib.Path('/path/to/my/file')
Then p.parents[0]
is the directory containing file
:
>>> p.parents[0]PosixPath('/path/to/my')
...and p.parents[1]
will be the next directory up:
>>> p.parents[1]PosixPath('/path/to')
Etc.
p.parent
is another way to ask for p.parents[0]
. You can convert a Path
into a string and get pretty much what you would expect:
>>> str(p.parent)'/path/to/my'
And also on any Path
you can use the .absolute()
method to get an absolute path:
>>> os.chdir('/etc')>>> p = pathlib.Path('../relative/path')>>> str(p.parent)'../relative'>>> str(p.parent.absolute())'/etc/../relative'
Note that os.path.dirname
and pathlib
treat paths with a trailing slash differently. The pathlib
parent of some/path/
is some
:
>>> p = pathlib.Path('some/path/')>>> p.parentPosixPath('some')
While os.path.dirname
on some/path/
returns some/path
:
>>> os.path.dirname('some/path/')'some/path'
I came here looking for something very similar. My solution, based on the above by @larsks, and assuming you want to preserve the entire path except the filename, is to do:
>>> import pathlib>>> p = pathlib.Path('/path/to/my/file')>>> pathlib.Path('/'.join(list(p.parts)[1:-1])+'/')
Essentially, list(p.parts)[1:-1]
creates a list of Path elements, starting from the second to n-1th, and you join them with a '/' and make a path of the resulting string. Edit The final +'/' adds in the trailing slash - adjust as required.